Math Problem Statement

You selected the correct​ choice, and your second answer is​ correct, but your first answer is incorrect. Use the derivative of the function to determine whether the function is increasing or decreasing in the intervals around any critical values found in the previous step in order to distinguish minima and maxima. Note that if a function f is increasing on both intervals around a critical value​ c, it is also increasing at the critical​ point, even if f prime​(c)equals​0, and the same holds for decreasing intervals.

Solution

It seems like you're working with a function and its derivative to determine the intervals of increase and decrease around its critical points, as well as distinguishing between minima and maxima. Here's how we can proceed:

Step-by-step approach:

  1. Find the derivative: Compute f(x)f'(x) of the function f(x)f(x).
  2. Locate the critical points: Critical points occur where f(x)=0f'(x) = 0 or f(x)f'(x) is undefined. You should have already identified these points.
  3. Determine intervals around critical points: To find out whether the function is increasing or decreasing, examine the sign of the derivative on the intervals that surround each critical point.
    • If f(x)>0f'(x) > 0, the function is increasing.
    • If f(x)<0f'(x) < 0, the function is decreasing.
  4. Classify critical points:
    • If the derivative changes sign from positive to negative at cc, then f(x)f(x) has a local maximum at cc.
    • If the derivative changes sign from negative to positive at cc, then f(x)f(x) has a local minimum at cc.
    • If the derivative does not change signs, cc is neither a local minimum nor maximum.

Example:

Let’s say we have a function f(x)=x33x2+2f(x) = x^3 - 3x^2 + 2.

  1. Find the derivative: f(x)=3x26xf'(x) = 3x^2 - 6x

  2. Find critical points: Set f(x)=0f'(x) = 0: 3x26x=0    3x(x2)=03x^2 - 6x = 0 \implies 3x(x - 2) = 0 So, x=0x = 0 and x=2x = 2 are the critical points.

  3. Determine intervals around the critical points: The critical points divide the number line into three intervals: (,0)(-\infty, 0), (0,2)(0, 2), and (2,)(2, \infty).

    • For (,0)(-\infty, 0), choose x=1x = -1: f(1)=3(1)26(1)=3+6=9f'(-1) = 3(-1)^2 - 6(-1) = 3 + 6 = 9. Since f(x)>0f'(x) > 0, the function is increasing in this interval.
    • For (0,2)(0, 2), choose x=1x = 1: f(1)=3(1)26(1)=36=3f'(1) = 3(1)^2 - 6(1) = 3 - 6 = -3. Since f(x)<0f'(x) < 0, the function is decreasing in this interval.
    • For (2,)(2, \infty), choose x=3x = 3: f(3)=3(3)26(3)=2718=9f'(3) = 3(3)^2 - 6(3) = 27 - 18 = 9. Since f(x)>0f'(x) > 0, the function is increasing in this interval.
  4. Classify critical points:

    • At x=0x = 0, the function changes from increasing to decreasing, so it has a local maximum.
    • At x=2x = 2, the function changes from decreasing to increasing, so it has a local minimum.

Conclusion:

  • The function is increasing on (,0)(-\infty, 0) and (2,)(2, \infty).
  • The function is decreasing on (0,2)(0, 2).
  • x=0x = 0 is a local maximum and x=2x = 2 is a local minimum.

Would you like to go through another example or discuss any specific part of this process in more detail?

Here are a few related questions:

  1. How do you apply the second derivative test to classify critical points?
  2. What happens when the derivative is undefined at a critical point?
  3. How can you determine if a function is concave up or concave down?
  4. What are inflection points, and how do they relate to the concavity of a function?
  5. How can you use a sign chart to analyze a function’s behavior?

Tip: When finding critical points, always double-check your derivative for errors before classifying intervals. It's easy to overlook mistakes in the process!

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Math Problem Analysis

Mathematical Concepts

Derivatives
Critical Points
Intervals of Increase and Decrease
Local Minima and Maxima

Formulas

f'(x) = 0 for critical points
Test for increasing/decreasing intervals (f'(x) > 0 for increasing, f'(x) < 0 for decreasing)

Theorems

First Derivative Test for Local Extrema

Suitable Grade Level

Grades 11-12